Question

In a theater, there are $20$ seats in the first row. Each row has $3$ more seats than the row ahead of it. There are $35$ rows in the theater. Find the total number of seats in the theater.

Answer 1

Compute the total number of seats in the theater:

It is given that the first row contain $20$ seats and each row has $3$ more seats than the row ahead it.

An arithmetic sequence with the first term $20$ and the common difference $3$ is obtained from the given details.

The total sum of an arithmetic sequence is calculated by $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$.

Where, $\mathrm{n}$ is the general term, $a$ is the first term and $d$ is the common difference.

It is given that the total number of rows in the theater is $35$.

Substitute $35$ for $\mathrm{n}$, $20$ for $a$ and $3$ for $d$ in $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$.

$$\begin{gathered} {\mathrm{S}}_{\mathrm{n}}=\frac{35}{2}\left(2\times 20+\left(35-1\right)3\right) \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{35}{2}\left(40+34\times 3\right) \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{35}{2}\left(142\right) \\ \therefore {\mathrm{S}}_{\mathrm{n}}=2485 \end{gathered}$$

Hence, the total number of seats in the theatre is $2485$.